Algebraic Theory of (arxiv: 1205.2882) 1 (in collaboration with T.R. Govindarajan, A. Queiroz and A.F. Reyes-Lega) 1 Physics Department, Syracuse University, Syracuse, N.Y. and The Institute of Mathematical Sciences, Chennai
Outline Introduction 1 Introduction 2 3 4
Introduction In studies of foundations of quantum theory, it is of interest to study mixed states and their origins. Focus has been on separable states and entropy created by partial tracing. But this method is not so good for identical particles as we will show. A much more universal construction is based on restrictions of states to subalgebras and the GNS construction. This talk will explain this approach
WHAT THEY DO Separable state Consider a bipartite system with Hilbert space H A H B. A vector state ψ = ψ ij i j i,j is said to be separable if it can be brought to the form ψ = v w. Otherwise, it is said to be entangled.
Singular value decomposition (SVD) A : m n complex matrix. A can always be written in the form A = UDV U : m m, unitary (columns of U are eigenvectors of AA ). D : m n, diagonal, positive (eigenvalues of A A.) V : n n unitary (columns of V are eigenvectors of A A).
Schmidt decomposition ψ = i,j A ij i j = i,j (UDV ) ij i j = U ik D kl V }{{} i,j,k,l =λ k δ kl lj i j = k λ k ( i ) U ik i j V jk j = k λ k k A k B Schmidt rank The Schmidt rank is the number of nonzero λ k s.
ψ separable precisely when Schmidt rank = 1. Reduced density matrix: ρ A = Tr B ρ. von Neumann entropy: S(ρ) = Trρ log ρ. We have S(ρ A ) = S(ρ B ). ψ separable precisely when S(ρ A ) = 0.
WHAT WE DO Our main motivation There are certain situations where the use of partial trace may not be the best thing to do. An example of this is provided by the study of entanglement for systems of indistinguishable particles, where the notion of separability is more subtle. Idea Partial trace = Restriction Thus consider two distinguishable particles A and B in a pure state ψ = φ A χ B. Partial trace means restricting its density matrix to observables of subsystem A. They are of the form K A 1 B. But for identical fermions...
Identical Fermions For identical fermions a two particle state is a linear combination of vector states of the form ψ = 1 2 ( φ χ χ φ ) and observables are all symmetric combinations K L + L K. Partial tracing has no physical meaning. How do we study the mixture created by observing only the single particle observables? K 1 + 1 K or perhaps L L? We turn now to this problem.
Outline Introduction 1 Introduction 2 3 4
C*-algebras Introduction C*-algebras Observables in quantum field theory come from C -algebras. All finite-dimensional matrix algebras are C. Representations of C -algebras Given a state or density matrix on such an algebra, there is a way to recover the Hilbert space due to Gelfand, Naimark and Segal. We will explain it below. It is used widely in Quantum field theory. Statistical physics. Noncommutative geometry.
Let A be the C -algebra of observables and ω : A C a state. That is ω(α) is a complex number, ω(α α) 0, and ω(1 A ) = 1. Regard A as a vector space: α α. Introduce a scalar product: α β = ω(α β) This space can have a subspace N of vectors of 0 norm: N = {α A α α = 0}. N is a left ideal: βn = N. Proved easily using Schwartz inequality. The Hilbert space is: H ω = A/N. An element of this space is the equivalence class [α] = α + N for α in A. The representation π ω of A on this space is given by π ω (α) [β] = [αβ]
von Neumann entropy and the GNS construction Fact: H ω irreducible precisely when ω pure. The state ω corresponds to the density matrix ρ ω = [1] [1]. H ω = j H j, where each H j carries an irreducible representation. If P j s are projectors from H ω to H j, set [1 j ] = P j [1] P j [1]. Define µ j = P j [1]. Then, j µ2 j = 1. Finally, ρ ω = j µ2 j [1 j] [1 j ]. This state is mixed if its rank exceeds 1. It has entropy S = µ 2 j log µ 2 j. j
Outline Introduction 1 Introduction 2 3 4
A = M 2 (C). ω λ : A C, ω λ (α) = λα 11 + (1 λ)α 22, 0 λ 1. The null space N ωλ is generated by those α such that Explicit form of this condition: ω λ (α α) = 0. λ( α 11 2 + α 21 2 ) + (1 λ)( α 12 2 + α 22 2 ) = 0. (1) Each value of λ gives a GNS-representation. There are two inequivalent cases: 0 < λ < 1 and λ = 0 (or 1). Notation: ( ) ( ) 1 0 0 1 e 11 =, e 0 0 12 =, etc.. 0 0
Case 1: 0 < λ < 1. For 0 < λ < 1 the only solution to (1) is α = 0 N ωλ = {0}. GNS-space given by H ωλ = C 4. The e ij act on this space as: π ωλ (e ij ) [e kl ] = δ jk [e il ]. For instance, the matrix of π ωλ (e 11 ) is: π ωλ (e 11 ) = 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0. H ωλ = H (1) H (2), with H (l) (l = 1, 2) invariant and spanned by { [e kl ] } k=1,2.
Case 1: 0 < λ < 1. Decomposing ω λ into pure states (1 A = 1 2 ): [1 A ] = [e 11 ] + [e 22 ] The norms of the two components are λ and 1 λ, so [1 A ] = λ [χ 1 ] + 1 λ [χ 2 ], [χ i ] χ j ] = δ ij. It follows that ρ ωλ = λ [χ 1 ] [χ 1 ] + (1 λ) [χ 2 ] [χ 2 ], so that ω λ is not pure. It has von Neumann entropy S(ω λ ) = λ log λ (1 λ) log(1 λ).
Case 2: λ = 0. Introduction If we choose λ = 0, from (1) we see that N ωλ = C 2, since it is spanned by elements of the form ( ) α11 0 α =, α 21 0 that is, by linear combinations of e 11 and e 21. Accordingly, the GNS-space H ωλ = Â/N ωλ = C 2 is generated by [e 12 ] and [e 22 ]. In this case the representation of A is irreducible and given by 2 2 matrices π ωλ (e ij ): π ωλ (e ij ) [e k2 ] = δ jk [e i2 ]. The state ω λ is pure with zero entropy. A similar situation is found for λ = 1.
Outline Introduction 1 Introduction 2 3 4
IDENTICAL PARTICLES: FERMIONS One-particle space Single-particle space: H C d. Symmetry group: U(d). Algebra of observables: A, given by a -representation of CU(d) for the group algebra: α = dµ(g)α(g)u (1) (g), U(d)
IDENTICAL PARTICLES: FERMIONS Many-particle space Consider now a fermion system whose single-particle space is given by H = C d : Many particle space F is the Fock space: F = d H (k), where H (k) Λ k H. k=0 Λ 0 H = C, generated by the vacuum" Ω. Λ 1 H: 1-particle space, Λ 2 H: 2-particle space, and so on.. dim Λ k H = ( d k), so dim F = 2 d. This reflects the fact that F = C 2 C 2, an isomorphism often used in statistical physics models.
{ e 1, e 2,... e d }: Orthonormal basis for H. Given v i H (i = 1,..., k), put v 1... v k = 1 sgn(σ)v 1 v k. k! σ S k Then: { e i1... e ik } 1 i1 < <i k d: Orthonormal basis for H (k). A self-adjoint operator A acting on H (1) H can be made to act on H (k) by defining dγ k (A) follows: dγ k (A) = A 1 d 1 d +1 d A 1 d 1 d + +1 d 1 d A This operator preserves the symmetry of the states on which it acts, as well as the commutation relations of the self-adjoint operators acting on H, namely: dγ k ([A, B]) = [dγ k (A), dγ k (B)]
Introduction dγ(a) = k dγk (A) acts on the whole Fock space F ( it is the second quantized" form of A). At the group level, we may consider exponentials of such operators, of the form e idγ(a). These are unitary operators acting on F. Let U = e ia be a unitary operator acting on H. The global version of dγ is given by: Γ k U = U U. We then have, with Γ(U) = k Γk (U), Γ(e ia ) = e idγ(a).
Following the previous remarks, we see that operators of the form α k = dµ(g)α(g)g g ( k-fold product, g U(d) ), U(d) act properly on H (k). All of this can be conveniently expressed in terms of a coproduct: approach based on Hopf algebras (can easily include braid-group statistics). Here, the construction of the observable algebra corresponds to the following simple choice for the coproduct : (g) = g g, g U(d), (2) linearly extended to all of CU(d). This choice fixes the form of α k.
Introduction is a homomorphism from the single-particle algebra to the two-particle Hilbert space. So it makes sense to identify its image with observations of single-particle observables.
Introduction An example with fermions Consider the 2-fermion space H (2) for the case d = 3. Let the action of U(3) on H = C 3 be given by the defining representation. Call it U (1). (Hence U (1) (g) = g). We have U (1) (g) e i = 3 j=1 D(g) ij e j, for a fixed ONB { e 1, e 2, e 3 }. The action of CU(3) is then given by the 3-dimensional conjugate representation (3 3 = 6 3). 3 is the antisymmetric 3 A 3. The basis vectors of 3 are f k := ε ijk e i e j (k = 1, 2, 3).
Action of CU(3) on H (1) : α = U(3) dµ(g)α(g)d(3) (g). Basis: the 8 Gell-Mann matrices {T i } i plus 1 3. On 3, they become T j = T j (use and restrict to 3). This amounts to consider only the action of the operators α 2 = dµ(g)α(g)d (3) (g) D (3) (g) U(3) on the (invariant) subspace generated by the antisymmetric vectors f k = ε ijk e i e j (k = 1, 2, 3). Thus, one-particle observables acting on the two-particle (fermionic) sector are generated by the matrices { T j T j } j=1,...8 of 3, plus the identity.
Introduction Summarizing: The algebra of operators A acting on the 2-particle sector H (2) = Λ 2 C 3 is the matrix algebra generated by { T 1,..., T 8, 1 3 }.
If we now assume that we only have access to the observables pertaining to the states e 1 and e 2, then the relevant algebra of operators will be a subalgebra A 0 A, namely the one generated by { T 1, T 2, T 3, 1 2, 1 3 }. In general we expect that a general (2-particle) pure state, given by a state vector ψ = 3 ψ k f k = ψ 1 e 2 e 3 + ψ 2 e 3 e 1 + ψ 3 e 1 e 2 k=1 may become mixed when restricted to A 0.
In order to detect this entanglement we perform the GNS construction when the state ψ ψ is restricted to A 0. We put ω ψ,o = ω ψ A0. furnishes a representation π : A 0 B(H ωψ,o ), as explained before. In general, H ωψ,o will split as a sum of irreducibles of A 0. This reducibility reflects the fact that, when restricted to A 0, the original state ω ψ might become mixed. The entropy of this state can then be computed from the decomposition of H ωψ,o into irreducibles.
Example: ψ = cos θ f 1 + sin θ f 3 cos θ e 2 e 3 + sin θ e 1 e 2 Null states The number of null states in the GNS construction based on ω ψ,o depends on the specific value of θ: For θ = 0, there will be 3 null states. For 0 < θ < π/2 we find 2 null states. For θ = π/2 we will get 4 null states.
Introduction von Neumann entropy of ω ψ,0 : Dimensions S(θ) = cos 2 θ log 1 cos 2 θ + sin2 θ log HGNS θ C 2, θ = 0 = C 3 = C 2 C, θ (0, π/2) C, θ = π/2. 1 sin 2 θ
Introduction Partial trace vs. GNS In the previous example (two fermions with single-particle space C 3 ), we obtain a θ-dependent entropy, whereas partial trace entropy always gives log 2 (since Slater rank 1, independently of θ). If C 4 describes single-particles, there are pure states of Slater rank 1 with zero for GNS entropy, log 2 for partial trace entropy (next example). The former is more reasonable, the state being least entangled.
Two Fermions, H (1) = C 4. Using creation/annihilation operators Algebras Fermionic creation/annihilation operators: a ( ) σ, b ( ) σ. a left, b : right. σ = 1, 2 spin up/down. Basis for H (2) : a 1 a 2 Ω, b 1 b 2 Ω, a σb σ Ω (σ, σ {1, 2}). A = M 6 (C) (4 4 = 10 6). A 0 : one-particle observables, left location. State ψ θ = ( ) cos θa 1 b 2 + sin θa 2 b 1 Ω.
Basis for A 0 : A 0 is the six-dimensional algebra generated by 1 A, T 1 := 1 2 (a 1 a 2 + a 2 a 1), T 2 := i 2 (a 1 a 2 a 2 a 1), T 3 := 1 2 (a 1 a 1 a 2 a 2), n 12 := (a 1 a 1a 2 a 2), N a := (a 1 a 1 + a 2 a 2). Entropy S(θ) = cos 2 θ log cos 2 θ sin 2 θ log sin 2 θ. Dimensions H θ = { C 2, θ = 0, π/2 C 4 = C 2 C 2, θ (0, π/2).
Remarks The significant aspect of this example is the fact that for the values of θ for which the Slater rank of ψ θ is one (θ = 0 and π 2 ), we obtain exactly zero for the entropy. In previous treatments of entanglement for identical particles, the minimum value for the von Neumann entropy of the reduced density matrix (obtained by partial trace) has been found to be log 2. This has been a source for controversy. Different entanglement criteria: non-identical particles, bosons, fermions..
Introduction An example with bosons. H (1) = C 3, with orthonormal basis{ e 1, e 2, e 3 }. The two-boson space H (2) : symmetrized vectors in H (1) H (1) (3 3 = 6 3). A 0 : subalgebra of one-particle observables pertaining only to the one-particle vectors e 1 and e 2. ψ (θ,φ) = sin θ cos φ e 1 e 2 + sin θ sin φ e 1 e 3 + cos θ e 3 e 3.
Dimensions The 6 SU(3) representation in 3 3 = 6 3 decomposes as 6 = 3 2 1 with respect to A 0. From this we can read off the decomposition of H (θ,φ) into irreducibles, depending on the coefficients of ψ. Entropy S(θ, φ) = sin 2 θ[cos 2 φ log(sin θ cos φ) 2 + sin 2 φ log(sin θ sin φ) 2 ] cos 2 θ log(cos θ) 2.
S(θ, φ) Introduction Figure: The two-boson entropy as a function of x and y which represent the (θ, φ)-sphere via stereographic projection. Darker regions correspond to lower values of the entropy.
CONCLUSIONS GNS-based approach: generalizes partial trace. Applications to entanglement of indistinguishable particles. Applications to quantum phase transitions? Applications to black hole physics?
3 3 = 6 3 in detail: u v in 3 3, with u = 3 i=1 ui e i and v = 3 j=1 v j e j. Decomposition into irreducible components: 3 i,j=1 u i v j e i e j = i j + i<j 1 2 (ui v j + u j v i )( e i e j + e j e i ) 1 2 (u i v j u j v i ) e i e j. First term: symmetric, 6-dimensional irreducible representation. Second term: antisymmetric, 3-dimensional complex irreducible representation, with basis vectors e k := ε ijk e i e j (k = 1, 2, 3), as stated above.
T 1 = T 4 = T 7 = 0 1 2 0 1 2 0 0 0 0 0 0 0 1 2 0 0 0 1 2 0 0 0 0 0 0 0 i 2 0 i 2 0, T 2 =, T 5 =, T 8 = 0 i 2 0 i 2 0 0 0 0 0 0 0 i 2 0 0 0 i 2 0 0 3, T 3 = 6 0 0 0 3 6 0 0 0 3 3, T 6 =, 1 3 = 1 2 0 0 0 1 2 0 0 0 0 0 0 0 0 0 1 2 0 1 2 0,, 1 0 0 0 1 0 0 0 1.